Given a circle, but not its center, construct an inscribed equilateral triangle in as few steps as possible.
I have a construction that works, but I am really having trouble understanding why the construction works. I will attach a photo of my construction below.
2 Answers
$\begingroup$$\triangle BDF$ is an equilateral triangle, so angle $\angle BDF$ has measure $60$. So $m\angle BGH=m\angle BDH=60$, since these two angles are subtended by the same arc $BH$. And $\triangle BED$ is also equilateral, so the remaining angle $\angle HDG$ on the line $EG$ has measure $60$. But $\angle HBG$ has the same measure as $\angle HDG$, since they are subtended by the same arc $HG$. This establishes that two of the angles in $\triangle BGH$ are $60$.
$\endgroup$ $\begingroup$Here is an alternative method, which requires identifying a diameter but not the center.
Select any point $A$ on the circle. Center the compasses there and draw an arc through two point $B,C$ on the circle.
Bisect $\angle BAC$, identifying point $D$ as the angle-interior point where the bisector intersects the circle. Center the compasses on each endpoint of $AD$ and draw an arc through the other endpoint, the two arcs intersecting at point $E$ (either of two choices).
Draw $AE$, which intersects the circle at point $F$ such that chord $DF$ measures one side of the triangle, and copy the chord around the circle accordingly.
Also $AF$ measures one side of an inscribed hexagon, so this polygon is obtainable too.
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